Dynamics: Normal and Tangential projectile motion problem

AI Thread Summary
The discussion focuses on a projectile motion problem involving a cannon firing at 100 m/s at a 20-degree angle. For part (a), the radius of curvature at the highest altitude is calculated using the horizontal component of velocity, leading to a corrected value of approximately 900.12 meters. Part (b) involves determining the radius of curvature 0.5 seconds after firing by finding the velocity vector at that time. Participants emphasize the importance of using the correct velocity components and the distinction between total acceleration and normal acceleration. The conversation highlights the application of projectile motion equations and vector analysis in solving the problem.
ajt49
Messages
2
Reaction score
0

Homework Statement



A canon fires a projectile at a speed of 100 m/s at angle of 20 degrees. (a) what is the radius of curvature of the projectile at its highest altitude? (b) What is the radius of curvature of the path .5 seconds after firing?

Homework Equations



Velocity vector = s'ur

Acceleration vector = v' =
atut + anun

where,

at= v'= dv/dt

&

an= [v^(2)]/p

The Attempt at a Solution



Wrote out the given which was,

Vo = 100 m/s at an angle 20 degrees above the horizontal x+ axis

Vox= Vo*cos(20)

Voy= Vo*sin(20)

ax = 0 = at = 0

ay= g or -9.81 acting in the negative j direction.

(a) an = -9.81 ==> -9.81 = [v^2]/p ==> p = [100^2]/9.81 = 1019.37 meters
 
Physics news on Phys.org
hey! welcome to physicsforums :)

Uh, your answer for part a is almost right. But v is not 100m/s. Because 100m/s is the initial velocity. So what velocity does the object have when at the highest altitude?
 
Thanks for the welcome and help!

Well at the highest altitude I guess the y component of the velocity is zero, so I would only have to calculate the x component of velocity. Therefore, an= [v^2]/p is actually

-9.81=[(100m/s*cos 20)^2]/p .. solve for row and p= 900.12 meters? Would I use the equations (Vosin theta)t and (Vocos theta)t at time .5 seconds? the are vectors so I could find my speed by Pythagorean theorem.
 
yes, I agree with your answer of 900.12m for part a. And for part b, yes, that sounds like a good plan. find the vector velocity at that time, and then you can find both an and |v|. (remember that an is not the total acceleration, but the acceleration in the direction perpendicular to the velocity).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top